Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
	Scanner scn = new Scanner(System.in);
	int n =scn.nextInt();
	double[]p = new double[n+1];
	for(int i=1;i<=n;i++)
	{
		p[i]=scn.nextDouble();
	}
	double[][]dp = new double[n+1][n+1];
	dp[0][0]=1;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<=i;j++)
		{
			if(j!=0)
			dp[i][j]=((1-p[i])*(dp[i-1][j]))+(p[i]*dp[i-1][j-1]);
			else {
				dp[i][j]=((1-p[i])*(dp[i-1][j]));
			}
			
		}
	}
	double ans=0;
	for(int j=(n+1)/2;j<=n;j++)
	{
		ans+=dp[n][j];
	}
	System.out.println(ans);
}
}